In p-adic metric, what is the distance between 0.9999... and 1?

Question

In 10-adic (though 10 is not a prime number) metric, we know that

$\Vert1-0.9999...\Vert=\Vert\lim_{n\rightarrow \infty}\frac{1}{10^n}\Vert=\lim_{n\rightarrow \infty}10^n\rightarrow\infty$.

However, from another point of view: $0.9999...=3\times\frac{1}{3}=1 $, we then should have

$\Vert1-0.9999...\Vert=\Vert1-1\Vert=\Vert0\Vert=0$

Which one should be right? Or, in 10-adic system, we should have

$0=\lim_{n\rightarrow \infty}10^n\rightarrow\infty$

or say, $0$ is infinitely close to $\infty$ in the 10-adic system?

Answer 1

The following series diverges with respect to the 10-adic absolute value (not measure) because of the first argument you give. $$0.9999... = \sum_{n=1}^\infty 9 *10^{-n}$$

This is not a representation of $1$ in the 10-adics just like

$$\frac{-1}{9} = ...1111 = \sum_{n=0}^\infty 10^n$$

is not a representation of $-1/9$ in the reals, but is perfectly fine in the context of 10-adics. A simple way to think of it is if a p-adic number has infinitely many digits to the right, it doesn't converge to represent a p-adic number just like a real number with infinitely many digits to the left doesn't converge to represent a real number.

Answer 2

The error in your first line is that you are not precise enough about what limits are taken with respect to which metric. If the expression $\lim_{n\rightarrow \infty}\frac{1}{10^n}$ is supposed to mean the limit with respect to $10$-adic value, then the first equality in that line is wrong; if it is supposed to mean the limit with respect to the usual Euclidean metric, then the second equality in that line is wrong.

The first equality i.e.

$\Vert1-0.9999...\Vert=\Vert\lim_{n\rightarrow \infty}\frac{1}{10^n}\Vert$

is true if the limit is understood to be with respect to usual (Euclidean) metric on $\mathbb Q$ just because both sides, and actually the expressions that sit inside the $\Vert \cdot \Vert$, are both $0$. (It is not true if the limit is meant w.r.t the $10$-adic topology, i.e. if that's what the limit is supposed to mean, your first mistake is already here.) Of course we would also have, and could write the same insight as,

$\Vert 0\Vert=\Vert\lim_{n\rightarrow \infty}\frac{1}{n}\Vert$

or

$\Vert0\Vert=\Vert\lim_{n\rightarrow \infty}(-1^n)\frac{\lfloor 3\sin(n)\rfloor}{7n}\Vert$

etc.

But then the following equality

$\Vert\lim_{n\rightarrow \infty}\frac{1}{10^n}\Vert\stackrel{?}=\lim_{n\rightarrow \infty}10^n$

is wrong, because even though it is true (for $\Vert \cdot \Vert$ meaning the $10$-adic value) that $\Vert \frac{1}{10^n}\Vert=10^n$ for each $n$, the map

$$x \mapsto \Vert x \Vert$$

is not continuous at $0$ when viewed as a map $(\mathbb Q, \text{Euclidean metric}) \rightarrow (\mathbb Q, \text{Euclidean metric})$, as your example, or mine above, all show. So you cannot just pull the limits out of that function.

(As an aside, in another answer I recently pointed out that that map would also not be continuous at $0$ when viewed as a map $(\mathbb Q, p\text{-adic metric}) \rightarrow (\mathbb Q, p\text{-adic metric})$. It would be continuous, by definition, when viewed as a map

$$(\mathbb Q, p\text{-adic metric}) \rightarrow (\mathbb Q, \text{Euclidean metric}).$$

But if you want to use that to make your second equality true, then, as said, you have to interpret the limit sitting inside the value signs in $\Vert\lim_{n\rightarrow \infty}\frac{1}{10^n}\Vert$ as a $10$-adic one, and then the first equality is wrong as pointed out at the beginning.)